编程垂直编码锯齿形矩阵

Question

我正在编写一个以zigzag模式打印nxn矩阵的动态代码.请帮助我获取以下输出的代码:

到目前为止,我在Rizier123的帮助下尝试过的代码是水平锯齿形图案:

#include <stdio.h>

int main() {

    int rows, columns;
    int rowCount, columnCount, count = 0;

    printf("Please enter rows and columns:\n>");
    scanf("%d %d", &rows, &columns);


    for(rowCount = 0; rowCount < rows; rowCount++) {

        for(columnCount = 1; columnCount <= columns; columnCount++) {

            if(count % 2 == 0)
                printf("%4d " , (columnCount+(rowCount*columns)));
            else
                printf("%4d " , ((rowCount+1)*columns)-columnCount+1);

        }
        count++;
        printf("\n");
    }


    return 0;

}

输入:

5 5

输出:

 1  2  3  4  5
10  9  8  7  6
11 12 13 14 15
20 19 18 17 16
21 22 23 24 25 

我想要相同的锯齿形图案输出,但垂直..

编辑

预期产量:

1 10 11 20 21 30
2  9 12 19 22 29
3  8 13 18 23 28
4  7 14 17 24 27
5  6 15 16 25 26

Answer 1

这应该适合你:

#include <stdio.h>

int main() {

    int rows, columns;
    int rowCount, columnCount;

    printf("Please enter rows and columns:\n>");
    scanf("%d %d", &rows, &columns);


    for(rowCount = 0; rowCount < rows; rowCount++) {

        for(columnCount = 0; columnCount < columns; columnCount++) {

          if(columnCount % 2 == 0)
                printf("%4d " , rows*(columnCount)+rowCount+1);
            else
                printf("%4d " , (rows*(columnCount+1))-rowCount);    

        }

        printf("\n");
    }


    return 0;

}

输入:

5 5

输出:

1  10  11  20  21
2   9  12  19  22
3   8  13  18  23
4   7  14  17  24
5   6  15  16  25